$\lim_{x\to -1}\dfrac{x+1}{\sqrt{x+5}-2}=$
Answer: Substituting $x=-1$ into $\dfrac{x+1}{\sqrt{x+5}-2}$ results in the indeterminate form $\dfrac{0}{0}$. This doesn't necessarily mean the limit doesn't exist, but it does mean we have to work a little before we find it. Since we have a rational expression with a square root on our hands, let's try to re-write it using the method of rationalization. $\begin{aligned} &\phantom{=}\dfrac{x+1}{\sqrt{x+5}-2} \\\\ &=\dfrac{x+1}{\sqrt{x+5}-2}\cdot\dfrac{\sqrt{x+5}+2}{\sqrt{x+5}+2} \gray{\text{Rationalize the denominator}} \\\\ &=\dfrac{(x+1)(\sqrt{x+5}+2)}{(x+5)-2^2} \\\\ &=\dfrac{\cancel{(x+1)}(\sqrt{x+5}+2)}{\cancel{(x+1)}} \gray{\text{Cancel out common factors}} \\\\ &=\sqrt{x+5}+2 \text{, for }x\neq -1 \end{aligned}$ This means that the two expressions have the same value for all $x$ -values (in their domains) except for $-1$. We can now use the following theorem: If $f(x)=g(x)$ for all $x$ -values in a given interval except for $x=c$, then $\lim_{x\to c}f(x)=\lim_{x\to c}g(x)$. In our case, $\dfrac{x+1}{\sqrt{x+5}-2}=\sqrt{x+5}+2$ for all $x$ -values in the interval $(-1.5,-0.5)$ except for $x=-1$. Therefore, $\lim_{x\to -1}\dfrac{x+1}{\sqrt{x+5}-2}=\lim_{x\to -1}\sqrt{x+5}+2=4$. (The last limit was found using direct substitution.) [I want to see how this looks graphically!] In conclusion, $\lim_{x\to -1}\dfrac{x+1}{\sqrt{x+5}-2}=4$.